TITLE: Proving an integral identity: $\int\nolimits_{-\infty}^\infty x f(x)f(t-x) dx =\frac{t}{2} \int_{-\infty}^\infty f(x)f(t-x) dx $
QUESTION [6 upvotes]: Let $f$ be a nonnegative (probably not needed) function on $\mathbb{R}$ such that for all $t$, $xf(x)f(t-x)$ and $f(x)f(t-x)$ are both integrable in $x$. 
Is it true that $$ \int\nolimits_{-\infty}^\infty x f(x)f(t-x) dx =\frac{t}{2} \int_{-\infty}^\infty f(x)f(t-x) dx $$
for all $t$?
I found that this is true if $f(x)=e^{-|x|} $ or $\frac{1}{1+x^2}$.

REPLY [9 votes]: It is true.  Notice that $$\int_{-\infty}^\infty xf(x)f(t-x)dx = \int_{-\infty}^\infty(t-x)f(t-x) f(x)dx.$$  This follows from the substitution $x=t-u$, or the fact that convolution is symmetric.
From here, use linearity of the integral to split up $(t-x)$, and you will find your equation.
Hope that helps